Question: $f(x, y) = \cos(x + y) - \cos(x - y)$ Is $f$ harmonic? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
A scalar field $f$ is harmonic if its Laplacian is zero. The Laplacian of a scalar field $f$ is the sum of each of its second partial derivatives. $\Delta f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2}$ [What does that triangle mean?] Let's find the second partial derivatives of $f$ ! $\begin{aligned} f_{xx} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ -\sin(x + y) + \sin(x - y) \right] \\ \\ &= -\cos(x + y) + \cos(x - y) \\ \\ f_{yy} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ -\sin(x + y) - \sin(x - y) \right] \\ \\ &= -\cos(x + y) + \cos(x - y) \end{aligned}$ The Laplacian: $\Delta f = -2\cos(x + y) + 2\cos(x - y) \neq 0$ Because the Laplacian of $f$ is nonzero, $f$ is not harmonic.